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Electric Potential in a Uniform Electric Field

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Electric Potential in a Uniform Electric Field
670
CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
or
KE i + PE i= KEf + PEf ,
(19.16)
where i and f stand for initial and final conditions. As we have found many times before, considering energy can give us insights and facilitate problem
solving.
Example 19.3 Electrical Potential Energy Converted to Kinetic Energy
Calculate the final speed of a free electron accelerated from rest through a potential difference of 100 V. (Assume that this numerical value is
accurate to three significant figures.)
Strategy
We have a system with only conservative forces. Assuming the electron is accelerated in a vacuum, and neglecting the gravitational force (we
will check on this assumption later), all of the electrical potential energy is converted into kinetic energy. We can identify the initial and final forms
of energy to be KE i = 0, KE f = ½mv 2, PE i = qV, and PE f = 0.
Solution
Conservation of energy states that
KE i + PE i= KE f + PE f .
(19.17)
2
qV = mv .
2
(19.18)
2qV
m .
(19.19)
Entering the forms identified above, we obtain
We solve this for
v:
v=
Entering values for
q, V, and m gives
v =
2⎛⎝–1.60×10 –19 C⎞⎠(–100 J/C)
(19.20)
9.11×10 –31 kg
= 5.93×10 6 m/s.
Discussion
Note that both the charge and the initial voltage are negative, as in Figure 19.4. From the discussions in Electric Charge and Electric Field, we
know that electrostatic forces on small particles are generally very large compared with the gravitational force. The large final speed confirms
that the gravitational force is indeed negligible here. The large speed also indicates how easy it is to accelerate electrons with small voltages
because of their very small mass. Voltages much higher than the 100 V in this problem are typically used in electron guns. Those higher voltages
produce electron speeds so great that relativistic effects must be taken into account. That is why a low voltage is considered (accurately) in this
example.
19.2 Electric Potential in a Uniform Electric Field
In the previous section, we explored the relationship between voltage and energy. In this section, we will explore the relationship between voltage and
electric field. For example, a uniform electric field E is produced by placing a potential difference (or voltage) ΔV across two parallel metal plates,
labeled A and B. (See Figure 19.5.) Examining this will tell us what voltage is needed to produce a certain electric field strength; it will also reveal a
more fundamental relationship between electric potential and electric field. From a physicist’s point of view, either ΔV or E can be used to describe
any charge distribution.
direction, while
ΔV is most closely tied to energy, whereas E is most closely related to force. ΔV is a scalar quantity and has no
E is a vector quantity, having both magnitude and direction. (Note that the magnitude of the electric field strength, a scalar quantity,
is represented by E below.) The relationship between ΔV and E is revealed by calculating the work done by the force in moving a charge from
point A to point B. But, as noted in Electric Potential Energy: Potential Difference, this is complex for arbitrary charge distributions, requiring
calculus. We therefore look at a uniform electric field as an interesting special case.
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CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
Figure 19.5 The relationship between
V
and
E
for parallel conducting plates is
E = V / d . (Note that ΔV = V AB
plate A at higher potential to plate B at lower potential, a minus sign needs to be included as follows:
The work done by the electric field in Figure 19.5 to move a positive charge
in magnitude. For a charge that is moved from
–ΔV = V A – V B = V AB . See the text for details.)
q from A, the positive plate, higher potential, to B, the negative plate,
lower potential, is
W = –ΔPE = – qΔV.
(19.21)
–ΔV = – (V B – V A) = V A – V B = V AB.
(19.22)
W = qV AB.
(19.23)
The potential difference between points A and B is
Entering this into the expression for work yields
Work is
W = Fd cos θ ; here cos θ = 1 , since the path is parallel to the field, and so W = Fd . Since F = qE , we see that W = qEd .
Substituting this expression for work into the previous equation gives
qEd = qV AB.
(19.24)
The charge cancels, and so the voltage between points A and B is seen to be
V AB = Ed⎫
V ⎬(uniform E - field only),
E = AB ⎭
d
(19.25)
where d is the distance from A to B, or the distance between the plates in Figure 19.5. Note that the above equation implies the units for electric
field are volts per meter. We already know the units for electric field are newtons per coulomb; thus the following relation among units is valid:
1 N / C = 1 V / m.
Voltage between Points A and B
where
V AB = Ed⎫
V ⎬(uniform E - field only),
E = AB ⎭
d
(19.26)
(19.27)
d is the distance from A to B, or the distance between the plates.
Example 19.4 What Is the Highest Voltage Possible between Two Plates?
6
Dry air will support a maximum electric field strength of about 3.0×10 V/m . Above that value, the field creates enough ionization in the air to
make the air a conductor. This allows a discharge or spark that reduces the field. What, then, is the maximum voltage between two parallel
conducting plates separated by 2.5 cm of dry air?
671
672
CHAPTER 19 | ELECTRIC POTENTIAL AND ELECTRIC FIELD
Strategy
We are given the maximum electric field
E between the plates and the distance d between them. The equation V AB = Ed can thus be used
to calculate the maximum voltage.
Solution
The potential difference or voltage between the plates is
Entering the given values for
V AB = Ed.
(19.28)
V AB = (3.0×10 6 V/m)(0.025 m) = 7.5×10 4 V
(19.29)
V AB = 75 kV.
(19.30)
E and d gives
or
(The answer is quoted to only two digits, since the maximum field strength is approximate.)
Discussion
One of the implications of this result is that it takes about 75 kV to make a spark jump across a 2.5 cm (1 in.) gap, or 150 kV for a 5 cm spark.
This limits the voltages that can exist between conductors, perhaps on a power transmission line. A smaller voltage will cause a spark if there are
points on the surface, since points create greater fields than smooth surfaces. Humid air breaks down at a lower field strength, meaning that a
smaller voltage will make a spark jump through humid air. The largest voltages can be built up, say with static electricity, on dry days.
Figure 19.6 A spark chamber is used to trace the paths of high-energy particles. Ionization created by the particles as they pass through the gas between the plates allows a
spark to jump. The sparks are perpendicular to the plates, following electric field lines between them. The potential difference between adjacent plates is not high enough to
cause sparks without the ionization produced by particles from accelerator experiments (or cosmic rays). (credit: Daderot, Wikimedia Commons)
Example 19.5 Field and Force inside an Electron Gun
(a) An electron gun has parallel plates separated by 4.00 cm and gives electrons 25.0 keV of energy. What is the electric field strength between
the plates? (b) What force would this field exert on a piece of plastic with a 0.500 µC charge that gets between the plates?
Strategy
Since the voltage and plate separation are given, the electric field strength can be calculated directly from the expression
electric field strength is known, the force on a charge is found using
equation in terms of the magnitudes,
F =qE.
E=
V AB
. Once the
d
F = q E . Since the electric field is in only one direction, we can write this
Solution for (a)
The expression for the magnitude of the electric field between two uniform metal plates is
E=
V AB
.
d
Since the electron is a single charge and is given 25.0 keV of energy, the potential difference must be 25.0 kV. Entering this value for
the plate separation of 0.0400 m, we obtain
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(19.31)
V AB and
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